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Divide the natural numbers in sets of consecutive numbers starting with {1,2} as the first set. The number of elements of the n-th set is equal to the sum of the n-1 final numbers in the (n-1)st set. The number of elements of the n-th set gives a(n).
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%I #7 Sep 10 2017 15:04:48

%S 2,2,7,30,158,985,7089,57890,529276,5358915,59543495,720476284,

%T 9431689530,132829627541,2002662076765,32185640519430,549301598198264,

%U 9921760117456135,189097075179752211,3792446896660586000,79840987298117599990,1760493769923493079769,40575189744903364314665

%N Divide the natural numbers in sets of consecutive numbers starting with {1,2} as the first set. The number of elements of the n-th set is equal to the sum of the n-1 final numbers in the (n-1)st set. The number of elements of the n-th set gives a(n).

%F a(n) = (1+a(1)+a(2)+...+a(n-1)-n/2)*(n-1).

%e The sets begin {1,2},{3,4},{5,6,...,9,10,11},{12,13,...,38,39,40,41},...

%o (PARI) a(n) = if (n==1, 2, (1+sum(k=1, (n-1), a(k)) - n/2)*(n-1)); \\ _Michel Marcus_, Sep 10 2017

%Y Cf. A067353.

%K easy,nonn

%O 1,1

%A _Floor van Lamoen_, Jan 17 2002

%E More terms from _Michel Marcus_, Sep 10 2017